Respuesta :

Answer:

a+2b-d=1, 3, 5, 7

Step-by-step explanation:

(ax^2+bx+3)(x+d)

ax^3+bx^2+3x+adx^2+bdx+3d

ax^3+bx^2+adx^2+3x+bdx+3d=x^3+6x^2+11x+12

ax^3=x^3, a=1

bx^2+adx^2=6x^2

x^2(b+ad)=6x^2

b+ad=6

b+(1)d=6

b+d=6

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3x+bdx=11x

x(3+bd)=11x

3+bd=11

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b=6-d

3+(6-d)d=11

3+6d-d^2=11

3-11+6d-d^2=0

-8+6d-d^2=0

d^2-6d+8=0

factor out,

(d-4)(d-2)=0

zero property,

d-4=0, d-2=0

d=0+4=4,

d=0+2=2

b=6-4=2,

b=6-2=4.

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a+2b-d=1+2(2)-2=1+4-2=5-2=3

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a+2(4)-4=1+8-4=9-4=5

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a+2(2)-4=1+4-4=5-4=1

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a+2(4)-2=1+8-2=9-2=7

Answer:

[tex]a+2b-d=1[/tex]

Step-by-step explanation:

We are given that:

[tex](ax^2+bx+3)(x+d)=x^3+6x^2+11x+12[/tex]

And we want to determine:

[tex]a+2b-d[/tex]

So, we will determine our unknowns first.

We can distribute our expression:

[tex]=(ax^2+bx+3)x+(ax^2+bx+3)d[/tex]

Distribute:

[tex]=ax^3+bx^2+3x+adx^2+bdx+3d[/tex]

Rearranging gives:

[tex]=(ax^3)+(bx^2+adx^2)+(bdx+3x)+3d[/tex]

Factoring out the variable yields:

[tex]=(a)x^3+(b+ad)x^2+(bd+3)x+d(3)[/tex]

Since we know that our expression equals:

[tex]x^3+6x^2+11x+12[/tex]

This means that each of the unknown terms in front of each variable corresponds with the coefficient of the resulting equation. Therefore:

[tex]\begin{aligned} a&=1\\ b+ad&=6\\bd+3&=11\\3d&=12\end{aligned}[/tex]

Solving the first and fourth equation yields that:

[tex]a=1\text{ and } d=4[/tex]

Then the second and third equations become:

[tex]b+(1)(4)=6\text{ and } b(4)+3=11[/tex]

And solving for b now yields that:

[tex]b=2\stackrel{\checkmark}{=}2[/tex]

Therefore, we know that:

[tex]a=1, b=2\text{ and } d=4[/tex]

For the equation:

[tex](x^2+2x+3)(x+4)=x^3+6x^2+11x+ 12[/tex]

Then the expression:

[tex]a+2b-d[/tex]

Can be evaluated as:

[tex]=(1)+2(2)-4[/tex]

Evaluate:

[tex]=1+4-4=1[/tex]

Hence, our final answer is 1.